Integrand size = 22, antiderivative size = 136 \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\frac {3}{8} \sqrt [3]{x^2+x^4}+\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{x^2+x^4}}{\sqrt {3}}\right )}{8\ 2^{2/3}}-\frac {\log \left (1+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{8\ 2^{2/3}}+\frac {\log \left (-1+2^{2/3} \sqrt [3]{x^2+x^4}-2 \sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{16\ 2^{2/3}} \]
3/8*(x^4+x^2)^(1/3)-1/16*3^(1/2)*arctan(-1/3*3^(1/2)+2/3*2^(2/3)*(x^4+x^2) ^(1/3)*3^(1/2))*2^(1/3)-1/16*ln(1+2^(2/3)*(x^4+x^2)^(1/3))*2^(1/3)+1/32*ln (-1+2^(2/3)*(x^4+x^2)^(1/3)-2*2^(1/3)*(x^4+x^2)^(2/3))*2^(1/3)
Time = 10.24 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\frac {1}{32} \left (12 \sqrt [3]{x^2+x^4}+2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1-2\ 2^{2/3} \sqrt [3]{x^2+x^4}}{\sqrt {3}}\right )-2 \sqrt [3]{2} \log \left (1+2^{2/3} \sqrt [3]{x^2+x^4}\right )+\sqrt [3]{2} \log \left (-1+2^{2/3} \sqrt [3]{x^2+x^4}-2 \sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )\right ) \]
(12*(x^2 + x^4)^(1/3) + 2*2^(1/3)*Sqrt[3]*ArcTan[(1 - 2*2^(2/3)*(x^2 + x^4 )^(1/3))/Sqrt[3]] - 2*2^(1/3)*Log[1 + 2^(2/3)*(x^2 + x^4)^(1/3)] + 2^(1/3) *Log[-1 + 2^(2/3)*(x^2 + x^4)^(1/3) - 2*2^(1/3)*(x^2 + x^4)^(2/3)])/32
Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1940, 1118, 27, 243, 60, 70, 16, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x \sqrt [3]{x^4+x^2}}{2 x^2+1} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt [3]{x^4+x^2}}{2 x^2+1}dx^2\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {1}{4} \int \frac {\sqrt [3]{x^4-1}}{2^{2/3} x^2}d\left (2 x^2+1\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt [3]{x^4-1}}{x^2}d\left (2 x^2+1\right )}{4\ 2^{2/3}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \frac {\sqrt [3]{2}}{\left (x^2\right )^{2/3}}dx^4}{8\ 2^{2/3}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 \sqrt [3]{2} \sqrt [3]{x^2}-\int \frac {1}{2^{2/3} \left (x^2\right )^{5/3}}dx^4}{8\ 2^{2/3}}\) |
\(\Big \downarrow \) 70 |
\(\displaystyle \frac {-\frac {3}{2} \int \frac {1}{2 x^2+2}d\left (\sqrt [3]{2} \sqrt [3]{x^2}\right )-\frac {3}{2} \int \frac {1}{\sqrt [3]{2} \sqrt [3]{x^2}-2 x^2}d\left (\sqrt [3]{2} \sqrt [3]{x^2}\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}}{8\ 2^{2/3}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {-\frac {3}{2} \int \frac {1}{\sqrt [3]{2} \sqrt [3]{x^2}-2 x^2}d\left (\sqrt [3]{2} \sqrt [3]{x^2}\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}-\frac {3}{2} \log \left (2 x^2+2\right )}{8\ 2^{2/3}}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {3 \int \frac {1}{-2 \sqrt [3]{2} \sqrt [3]{x^2}-2}d\left (2 \sqrt [3]{2} \sqrt [3]{x^2}-1\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}-\frac {3}{2} \log \left (2 x^2+2\right )}{8\ 2^{2/3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {-\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{2} \sqrt [3]{x^2}-1}{\sqrt {3}}\right )+\frac {\log \left (x^4\right )}{2}+3 \sqrt [3]{2} \sqrt [3]{x^2}-\frac {3}{2} \log \left (2 x^2+2\right )}{8\ 2^{2/3}}\) |
(3*2^(1/3)*(x^2)^(1/3) - Sqrt[3]*ArcTan[(-1 + 2*2^(1/3)*(x^2)^(1/3))/Sqrt[ 3]] + Log[x^4]/2 - (3*Log[2 + 2*x^2])/2)/(8*2^(2/3))
3.20.42.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) , x] + (Simp[3/(2*b*q) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 /3)], x] + Simp[3/(2*b*q^2) Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 17.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82
method | result | size |
pseudoelliptic | \(\frac {3 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{8}-\frac {2^{\frac {1}{3}} \ln \left (2 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+2^{\frac {1}{3}}\right )}{16}+\frac {2^{\frac {1}{3}} \ln \left (4 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}-2 \,2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )}{32}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, 2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{3}-\frac {\sqrt {3}}{3}\right )}{16}\) | \(111\) |
risch | \(\text {Expression too large to display}\) | \(1984\) |
3/8*(x^2*(x^2+1))^(1/3)-1/16*2^(1/3)*ln(2*(x^2*(x^2+1))^(1/3)+2^(1/3))+1/3 2*2^(1/3)*ln(4*(x^2*(x^2+1))^(2/3)-2*2^(1/3)*(x^2*(x^2+1))^(1/3)+2^(2/3))- 1/16*2^(1/3)*3^(1/2)*arctan(2/3*3^(1/2)*2^(2/3)*(x^2*(x^2+1))^(1/3)-1/3*3^ (1/2))
Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94 \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\frac {1}{16} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}}\right )}\right ) - \frac {1}{64} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} + 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}\right ) + \frac {1}{32} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} \]
1/16*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*4^(2/3)*(-1) ^(2/3)*(x^4 + x^2)^(1/3) - 4^(1/3))) - 1/64*4^(2/3)*(-1)^(1/3)*log(4^(2/3) *(-1)^(1/3)*(x^4 + x^2)^(1/3) + 4^(1/3)*(-1)^(2/3) + 4*(x^4 + x^2)^(2/3)) + 1/32*4^(2/3)*(-1)^(1/3)*log(-4^(2/3)*(-1)^(1/3) + 4*(x^4 + x^2)^(1/3)) + 3/8*(x^4 + x^2)^(1/3)
\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int \frac {x \sqrt [3]{x^{2} \left (x^{2} + 1\right )}}{2 x^{2} + 1}\, dx \]
\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int { \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{2 \, x^{2} + 1} \,d x } \]
\[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int { \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{2 \, x^{2} + 1} \,d x } \]
Timed out. \[ \int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx=\int \frac {x\,{\left (x^4+x^2\right )}^{1/3}}{2\,x^2+1} \,d x \]